WakeUp wrote: Prove that for all positive real numbers $a,b,c$ we have \[\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}\ge\sqrt{a^2+ac+c^2} \] After squaring both sides we have: $2b^2+2\cdot \sqrt{a^2-ab+b^2}\cdot \sqrt{b^2-bc+c^2} \ge ab+ac+bc$ Apply substitution $\frac{a}{b}=x$, and $\frac{c}{b}=y$, and square both sides, so it remains to prove: $4(x^2-x+1)(y^2-y+1) \ge (x+y+xy-2)^2$ $\Leftrightarrow 3(x+y-xy)^2 \ge 0$ With equality case: $a=c=2b$

WakeUp wrote: Prove that for all positive real numbers $a,b,c$ we have \[\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}\ge\sqrt{a^2+ac+c^2} \] Using Minkowski's inequality: $\sqrt{\left(\frac{a\sqrt{3}}{2}\right)^2+\left(\frac{a}{2}-b\right)^2} + \sqrt{\left(\frac{c\sqrt{3}}{2}\right)^2+\left(b-\frac{c}{2}\right)^2} \ge \sqrt{\left(\frac{a}{2}-\frac{c}{2}\right)^2+\frac{3}{4}(a+c)^2} = \sqrt{a^2+ca+c^2} $

WakeUp wrote: Prove that for all positive real numbers $a,b,c$ we have \[\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}\ge\sqrt{a^2+ac+c^2} \] Another proof: (using coordinate) Let 3 points : ${A(\frac{a\sqrt{3}}{2};\frac{a}{2}-b})$ ${B(\frac{c\sqrt{3}}{2};\frac{c}{2}-b})$ Then using the result : $ AB \le OA + OB $

WakeUp wrote: Prove that for all positive real numbers $a,b,c$ we have \[\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}\ge\sqrt{a^2+ac+c^2} \] After expanding both side we have $a^2b^2+b^2c^2+c^2a^2+2b^2ac\geq 2a^2bc+2c^2ab$ so it can be written as $b^2(a+c)^2+a^2c^2 \geq 2abc(a+c)$ which is true by A.M G.M and $b=\frac{ac}{a+c}$ will be equal

Let $ABCD$ be a convex quadrilateral. Construct $ABCD$ such that $\angle ADB=60, \angle BDC=60, AD=a, BD=b, CD=c$. By the Law of Cosines: $$\triangle ADC\to AC=\sqrt{a^2+ac+c^2}$$$$\triangle BDC\to BC=\sqrt{b^2-bc+c^2}$$$$\triangle ADB\to AB=\sqrt{a^2-ab+c^2}$$And by the triangle inequality in $\triangle ABC$, \[\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}\ge\sqrt{a^2+ac+c^2} \]We are done because the quadrilateral is clearly always constructible for any $a,b,c>0$. Equality occurs when $\triangle ABC$ is degenerate (or $a=2b=c$).

This is actually an AoPS Precalculus Writing Problem! Had no idea. Here's my writeup.

OlympusHero wrote: This is actually an AoPS Precalculus Writing Problem! O wait yea it was, just realized lol

Dehomogenize by setting $abc=1$. Square both sides twice: $$\Leftrightarrow2b^2+2\sqrt{(a^2-ab+b^2)(b^2-bc+c^2)}\ge ab+bc+ca$$$$\Leftrightarrow4\sum_{\text{cyc}}a^2b^2-4a+4b-4c-4ab^3-4b^3c+4b^4\ge(ab+bc+ca-2b^2)^2$$$$\Leftrightarrow\sum_{\text{cyc}}a^2b^2\ge2a-2b+2c$$$$\Leftrightarrow b^2(a+c)^2+a^2c^2\ge2(a+c)$$$$\Leftrightarrow (a+c)^2+a^4c^4\ge2a^2c^2(a+c)$$$$\Leftrightarrow x^2+y^2\ge2xy$$$$\Leftrightarrow(x-y)^2\ge0$$where $x=a+c$ and $y=a^2c^2$.

Although this inequality has a very beatiful geometric interpretation, I will use some algebra transformation and Minkowski inequality (which is very natural) to prove it. First note, that $a^2-ab+b^2=\frac{(a-b)^2}{2}+\frac{a^2}{2}+\frac{b^2}{2}$. Similarly, $b^2-bc+c^2=\frac{(c-b)^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}$. That`s why $\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}=\sqrt{\frac{(a-b)^2}{2}+\frac{a^2}{2}+\frac{b^2}{2}}+\sqrt{\frac{(c-b)^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}}$. Now, Minkowski inequality: $\sqrt{{x_1}^2+{x_2}^2+...+{x_n}^2}+\sqrt{{y_1}^2+{y_2}^2+...+{y_n}^2}\geq{\sqrt{({x_1+y_1})^2+({x_2+y_2})^2+...+({x_n+y_n})^2}}$. We set $n=3$ and $x_1={(a-b)^2}, y_1={b^2}, x_2=a^2, y_2=c^2, x_3=b^2, y_3=(c-b)^2$. We get: $\sqrt{(a-b)^2+a^2+b^2}+\sqrt{b^2+c^2+(c-b)^2}\geq{\sqrt{a^2+(a+c)^2+c^2}}$. That` s why: $\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}=\sqrt{\frac{(a-b)^2}{2}+\frac{a^2}{2}+\frac{b^2}{2}}+\sqrt{\frac{(c-b)^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}}\geq{\sqrt{\frac{a^2}{2}+\frac{(a+c)^2}{2}+\frac{c^2}{2}}}=\sqrt{\frac{2a^2+2ac+2c^2}{2}}=\sqrt{a^2+ac+c^2}$ - $Q.E.D.$